Question

when a polynomial 4x4-3x3+2x2-x+1 is divided by another polynomial x²+2 the remainder is ax+b find a and b​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm :\longmapsto\:f(x) = {4x}^{4} - {3x}^{3} + {2x}^{2} - x + 1$

and

$\rm :\longmapsto\:g(x) = {x}^{2} + 2$

Now, We use method of Long Division to get remainder.

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\: \: \: \: \: \: \: {4x}^{2} - 3x - 2 \: \: \: \: \:\:}}}\\ {{\sf{ {x}^{2} + 1}}}& {\sf{\: {4x}^{4} - {3x}^{3}+{2x}^{2} - x + 1\:}} \\{\sf{}}&\underline{\sf{\: \: \: \: \: { - 4x}^{4} \: \: \: \: \: - 4 {x}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}}\\{\sf{}}&{\sf{\: \: { - 3x}^{3} - {2x}^{2} - x + 1\:\:}}\\{\sf{}}&\underline{\sf{ \: { 3x}^{3} \: \: \: \: \: \: \: \: \: + 3x \:\:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: - {2x}^{2} + 2x + 1\:\:}}\\{\sf{}}&\underline{\sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: { 2x}^{3} \: \: \: \: \: \: \: \: \: +2 \:\:}}\\{\sf{}}&\underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: 2x \: + 3\:\: \: \: \: \: \: \: \: \: }}\end{array}\end{gathered}\end{gathered}\end{gathered}$

Hence,

We get

• Remainder = 2x + 3

But it is given that when f(x) is divided by g(x), the remainder is ax + b.

So,

$\rm :\longmapsto\:Remainder = ax + b = 2x + 3$

So, on comparing we get

$\rm :\longmapsto\:a = 2 \: \: \: and \: \: \: b = 3$

Additional Information :-

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: a {x}^{2} + bx + c, \: then$

$\red{ \boxed{ \rm{ \: \alpha + \beta = - \: \dfrac{b}{a} }}}$

and

$\red{ \boxed{ \rm{ \: \alpha \beta = \: \dfrac{c}{a} }}}$

$\rm :\longmapsto\: \alpha , \beta, \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then$

$\red{ \boxed{ \rm{ \: \alpha + \beta + \gamma = - \: \dfrac{b}{a} }}}$

$\red{ \boxed{ \rm{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a} }}}$

$\red{ \boxed{ \rm{ \: \alpha \beta \gamma = - \: \dfrac{d}{a} }}}$